# Prove that the integral of sup f_n is infinite

1. Aug 6, 2015

### QIsReluctant

1. The problem statement, all variables and given/known data
For n = 1, 2, ..., let fn be a Lebesgue integrable function [0,1] → [0, +∞) such that
(1) ∫01 fn dx = 1
and
(2) ∫1/n1 fn dx < 1/n

Let g(x) = supn ∈ ℕfn(x). Prove
01 g(x)dx = +∞

3. The attempt at a solution
Coffee, banging my head against a wall, etc.

There's not enough information about the behaviour of the fn themselves, apart from the integrals, to use any of the convergence theorems. Fatou's lemma seems like an obvious choice since it only deals in integrals, but choosing a positive and measurable function that leads to any results is daunting. This is another one of those problems where I get why the answer is what it is on a non-rigorous level but have no idea how to apply theory ...

I thought about turning the integral of g into an infinite sum of integrals from 1/(n-1) to 1/n, but since we can concentrate the support of fn in [0, 1/n] on as small an interval containing 0 as we want, we have no guarantee that any of those individual integrals is anything but positive. Aaaaargh.

2. Aug 7, 2015

### RUber

Using the fact that the integral from 0 to 1 is equal to 1 and the right side is shrinking to zero as n grows, you are essentially saying that:
$\int_0^{1/n}f dx =1-\int_{1/n}^1 f dx$
The integral of the sup function will be the max value the function takes multiplied by the length of the integral.

3. Aug 7, 2015

### QIsReluctant

Not necessarily so -- that would only be true if the sup function were constant. Or have I just misunderstood?

4. Aug 7, 2015

### RUber

Hmmm...that seems a bit ambiguous.
If we use your interpretation, then the problem is clearly more challenging.

I suppose if you define the smallest possible measureable interval, and try to determine the sup over that interval, you will force the integral into one that must be infinite.

Assume that f_n has the smallest sup possible by making it a constant function from 0 to 1/n and zero on the rest of the interval.

What is g(0)?
What is g(0+epsilon)?

In the end you should be able to show that the supremum function will be infinite for some measurable interval.

5. Aug 7, 2015

### QIsReluctant

Ach, I'm really not getting anywhere. Would you mind being more detailed?

The biggest problem for me is that we have no (apparent) guarantee that fn(x) will ever hold some minimum value anywhere. So g(ε) might be zero for all we know.

I can find plenty of lower bounds for sup f on [0, 1/n], but those are not really helpful. I've also tried searching for some kind of lower bound to μ(En), where En is the set where fn > n. Again, no luck -- all I can find are upper bounds for that measure.

6. Aug 7, 2015

### haruspex

I would look for sequences xi, ni, ai such that 0<xi+1<xi<1, ni+1>ni, $\Sigma a_i$ diverges, and $\int_{x_{i+1}}^{x_i}f_{n_i} > a_i>0$.

7. Aug 8, 2015

### haruspex

Are you in need of a stronger hint?
Given some xi between 0 and 1, how might you construct an ni, as a function of xi, such that you can put a useful lower bound on $\int _0^{x_i}f_{n_i}$?
Can you then show the existence of xi+1>0 such that you can put a useful lower bound on $\int _{x_{i+1}}^{x_i}f_{n_i}$?
What would that then give you as lower bound on $\int _{x_{i+1}}^{x_i}g$?

8. Aug 10, 2015

### QIsReluctant

I gave it a little bit of thought before going to bed last night, and the answer is now obvious

Since ∫[0,1/n] fn(x)dx > (n-1)/n, we have that
sup[0,1/n] fn(x) ≥ n - 1.

But since g(x) ≥ fn(x) for all x and n, we have that
[1/n, 1/n+1] g(x)dx ≥ (n - 1)[ 1/n - 1/(n+1) ].

We represent ∫[0,1] g(x)dx as an infinite sum of these integrals and get a sum that diverges by the limit rule; thus ∫[0,1] g(x)dx = ∞ QED.

Ach, this is false, in fact. Back to the drawing board ...

Last edited: Aug 10, 2015
9. Aug 10, 2015

### QIsReluctant

I think I've really got it this time, thanks to some help from someone who had the same question.

Suppose ∫01 g(x)dx < ∞. Then g is Lebesgue integrable since it is positive; thus the Dominated Convergence Theorem implies that ∫[0, 1/n] g(x)dx → 0. But this contradicts the hypothesis that says that ∫[0, 1/n] g(x)dx ≥ 1 - 1/n.

Thus ∫01 g(x)dx = ∞.

10. Aug 10, 2015

### NihilTico

The DCT implies that , $\lim_{n\to\infty}\int_{0}^{1}f_n d\mu=\int_{0}^{1}gd\mu$ for a sequence $f_n$ that converges pointwise to $g$. All you have is a pointwise supremum, $g=\sup_n f_n$, which is easily seen to be measurable. In any case, I can't see how the DCT would imply here that $\int_{[0,1/n]}gd\mu\to0$ unless I've been removed from measure theory for too long. Haruspex's method looks applicable here.

11. Aug 12, 2015

### davidmoore63@y

Trying to follow haruspex hint. Apologies no latex. We can try x(i)= 1/n. Then integral (0, x(i)) of f_n(x) fx is greater than 1-1/n. Next step, we know by the MVT there must be a x(i+1) such that 0<x(i+1)<x(i) and f_n(x(i+1))> n-1. But I don't see how this provides a bound for integral (x(I),x(i+1)) of f_n(x)fx. Need a stronger hint

12. Aug 12, 2015

### RUber

If you were to create a continuous function to represent the lower bound for g(x), it would be $g(x) = \frac{1}{x}$. What is $\int_0^1 \frac{1}{x} dx$?

Clearly, $\lim_{n\to \infty} f_n(x) = \delta(x)$. This is the standard formulation of the distributional Dirac delta function.
The problem is that the Dirac delta is only non-zero on a set of measure 0. If that non-zero component had a measure (like g(x) does), no matter how small it is, what would the integral need to be?
I know this is not the method for the proof, but it should make sense that it must be infinite.

13. Aug 12, 2015

### haruspex

Given $1>a_i>0$ choose $n_i>\frac 1{a_i}$.
$\lim x\rightarrow 0 \int_x^{a_i}f_{n_i} = \int_0^{a_i}f_{n_i} > \int_0^{\frac 1{n_i}}f_{n_i} > 1 - \frac 1{n_i} \geq 0.5$
So $\exists a_{i+1}>0$ such that .....
Can you continue from there?
It worries me that I don't seem to need $\int_{\frac 1n}^1 f_n < \frac 1n$, merely that $\int_{\frac 1n}^1 f_n < c$ for some constant c < 1.